It's Child's Play

In order to explain the solution, let us label the missing numbers as $n_1,\cdots,n_9$ . Thus, the problem statement can be expressed as :

$n_1+n_2+13 \times \frac{n_3}{n_4} + 12 \times n_5 + \frac{n_6 \times n_7}{n_8} - n_9 = 87$

With the condition that all the $n_i$ are distinct and lie between 1 and 9.

It can also be noticed that :

(i) If $n_1=a,n_2=b,n_3,\cdots,n_9$ is a solution, $n_1=b,n_2=a,n_3,\cdots,n_9$ is also a solution.

(ii) Similarly, $n_6$ and $n_7$ are also interchangeable.

Performing an exhaustive search would give 128 distinct solutions to the problem.

However, many of them lead to fractions during the division operations.

Eg :

$4 + 1 + 13 \times \frac{9}{ 6} + 12 \times 5 + \frac{7 \times 3}{ 2}- 8 = 87$

Restricting to solutions with a condition that $n_3$ divides $n_4$ leads to 5 sets of 4 solutions (resulting from the interchange of $n_1$ and $n_2$ AND $n_6$ and $n_7$ )

( 3,     5),    2,     1,     4,     (8,     9),     6,     7
( 6,     9),    3,     1,     2,     (7,     8),     4,     5
( 5,     7),    3,     1,     2,     (8,     9),     4,     6
( 5,     9),    4,     1,     2,     (3,     8),     6,     7
( 5,     6),     9,     3,     2,     (7,     8),     4,     1

Each of the above lines represents 4 valid solutions which can be generated by interchanging the numbers in either of the brackets.

Thus, there would be $\boxed{20}$ solutions (max. number of students)